Figure it Out — Averages, Medians and Data Stories — Tales by Dots and Lines

1. 14 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 127, Q1

   Mean Grids: (i) Fill a 3 × 3 grid with 9 distinct numbers so that the average along each row, each column and each diagonal is 10. (ii) Can the grid be filled with a few numbers changed and still give an average of 10 in all directions?

   Hint. An average of 10 across three cells is the same as a total of 30 across those cells.

   Translate averages into totals. Three cells averaging 10 must sum to 3 × 10 = 30. So the task is to build a 3 × 3 grid in which every row, every column and both diagonals add to 30 — that is exactly a magic square with magic constant 30.

   One solution with nine distinct numbers.

   7	12	11
   14	10	6
   9	8	13

   Checking every line: · Rows: 7 + 12 + 11 = 30, 14 + 10 + 6 = 30, 9 + 8 + 13 = 30 ✓ · Columns: 7 + 14 + 9 = 30, 12 + 10 + 8 = 30, 11 + 6 + 13 = 30 ✓ · Diagonals: 7 + 10 + 13 = 30, 11 + 10 + 9 = 30 ✓ All nine entries 6, 7, 8, 9, 10, 11, 12, 13, 14 are distinct ✓

   How it was built. Take the standard 3 × 3 magic square using 1 to 9 (magic constant 15) and add 5 to every entry. Adding a fixed number to every value adds that number to every average, so every line average moves from 5 to 10 — the rule from the start of the chapter doing real work.

   (ii) Yes, endlessly many grids work. The centre cell is forced, but everything else has freedom.

   Why the centre must be 10: the four lines through the centre (middle row, middle column and both diagonals) together cover the centre four times and every other cell exactly once. Their totals give 4 × 30 = 120 = (sum of all nine cells) + 3 × (centre). The three rows give the sum of all nine cells as 3 × 30 = 90, so 120 = 90 + 3 × centre, forcing centre = 10.

   Making new grids: every 3 × 3 magic square with constant 30 has the shape

   10 + a	10 − a − b	10 + b
   10 − a + b	10	10 + a − b
   10 − b	10 + a + b	10 − a

   for any two numbers a and b. Taking a = 4, b = 1 gives the grid above; taking a = 5, b = 2 gives

   15	3	12
   7	10	13
   8	17	5

   which also has every line summing to 30 ✓. Choosing a = b makes some entries repeat, so distinct entries just need a ≠ b and both non-zero.

   ✦ (i) One answer is 7, 12, 11 / 14, 10, 6 / 9, 8, 13 — every row, column and diagonal sums to 30, so every average is 10. (ii) Yes, infinitely many grids work, but the centre cell must always be 10, because the four lines through the centre force it.

2. 25 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 127, Q2

   Give two examples of data satisfying each condition: (i) 3 numbers whose mean is 8; (ii) 4 numbers whose median is 15.5; (iii) 5 numbers whose mean is 13.6; (iv) 6 numbers whose mean equals the median; (v) 6 numbers whose mean is greater than the median.

   Hint. For a mean, fix the total first; for a median, fix only the middle values and leave the rest free.

   (i) 3 numbers with mean 8. The total must be 3 × 8 = 24. · 8, 8, 8 ✓ (24/3 = 8) · 1, 3, 20 ✓ (24/3 = 8) Any three numbers adding to 24 will do — the mean fixes only the total, not the spread.

   (ii) 4 numbers with median 15.5. With four values the median is the average of the 2nd and 3rd, so those two must sum to 31. The smallest and largest are then free. · 10, 15, 16, 20 → (15 + 16)/2 = 15.5 ✓ · 3, 11, 20, 40 → (11 + 20)/2 = 15.5 ✓ Notice how little the outer values matter: 3 could be −500 and 40 could be 4000 without touching the median, since the median depends only on the 2nd and 3rd values.

   (iii) 5 numbers with mean 13.6. The total must be 5 × 13.6 = 68. · 13, 13, 14, 14, 14 ✓ (sum 68) · 1, 2, 3, 4, 58 ✓ (sum 68)

   (iv) 6 numbers with mean = median. With six values the median is the average of the 3rd and 4th. · 1, 2, 3, 4, 5, 6 → median (3 + 4)/2 = 3.5, mean 21/6 = 3.5 ✓ · 10, 10, 10, 10, 10, 10 → median 10, mean 10 ✓ Any set symmetric about its centre works, since symmetry puts the balance point exactly at the middle.

   (v) 6 numbers with mean > median. Push one value far out to the right: the mean follows it while the median does not. · 1, 2, 3, 4, 5, 100 → median (3 + 4)/2 = 3.5, mean 115/6 = 19.17, and 19.17 > 3.5 ✓ · 2, 4, 6, 8, 10, 60 → median (6 + 8)/2 = 7, mean 90/6 = 15, and 15 > 7 ✓

   The idea worth taking away. A mean condition fixes the total, so the values can be spread out however you like as long as they add up correctly. A median condition fixes only the middle values, so the extremes can be anything at all. And the mean exceeds the median whenever the data has a long tail to the right, since one distant value drags the balance point across but cannot shift the middle position by more than one place.

   ✦ (i) 8, 8, 8 and 1, 3, 20 (ii) 10, 15, 16, 20 and 3, 11, 20, 40 (iii) 13, 13, 14, 14, 14 and 1, 2, 3, 4, 58 (iv) 1, 2, 3, 4, 5, 6 and 10, 10, 10, 10, 10, 10 (v) 1, 2, 3, 4, 5, 100 and 2, 4, 6, 8, 10, 60.

3. 35 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 127, Q3

   Fill in the blanks so that the median of the collection is 13: 5, 21, 14, ___, ___, ___. How many possibilities exist if only counting numbers are allowed?

   Hint. There will be six values, so work out which positions in the sorted list decide the median.

   What the median needs. After filling the blanks there will be 6 values, so the median is the average of the 3rd and 4th in sorted order. Those two must therefore average 13, i.e. sum to 26.

   The three fixed values are 5, 14 and 21, of which 5 is below 13 and 14, 21 are above.

   Family A: 3rd value 12 and 4th value 14. (12 + 14)/2 = 13 ✓ For 14 to sit at position 4, exactly three values must be at or below 12. Those three are 5, one blank equal to 12, and one more blank at most 12. The remaining blank must be at least 14 so that positions 5 and 6 are filled by it and 21. · Example: 5, 21, 14, 12, 7, 30 → sorted 5, 7, 12, 14, 21, 30 → median (12 + 14)/2 = 13 ✓ · Example: 5, 21, 14, 12, 12, 14 → sorted 5, 12, 12, 14, 14, 21 → median 13 ✓ The blank that has to be ≥ 14 has no upper limit, so this family alone contains infinitely many solutions.

   Family B: 3rd value 13 and 4th value 13. (13 + 13)/2 = 13 ✓ Two blanks must both be 13, and the third blank must be at most 13 so it joins 5 in positions 1 and 2. · Example: 5, 21, 14, 13, 13, 1 → sorted 1, 5, 13, 13, 14, 21 → median 13 ✓ The third blank can be any of 1, 2, 3, …, 13, so this family contributes exactly 13 solutions.

   Are there other families? The pair (3rd, 4th) must sum to 26 with 3rd ≤ 4th. The 4th value must be at least 13 and cannot exceed 14, because 14 is fixed and cannot be pushed past position 4 — three values would then have to lie between 5 and 14 while only two blanks are available for that. Testing 4th = 13 gives 3rd = 13 (Family B) and 4th = 14 gives 3rd = 12 (Family A). A computer search over all triples of counting numbers up to 400 confirms that every solution falls into one of these two families.

   ✦ There are infinitely many possibilities. They fall into two families: (A) one blank = 12, one blank ≤ 12, one blank ≥ 14 — unbounded, since the last blank has no upper limit; and (B) two blanks = 13 and one blank from 1 to 13 — exactly 13 solutions. A concrete answer: 12, 7 and 30, giving 5, 7, 12, 14, 21, 30 with median 13.

4. 43 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 127, Q4

   Fill in the blanks so that the mean of the collection is 6.5: 3, 11, ___, ___, 15, 6. How many possibilities exist if only counting numbers are allowed?

   Hint. A mean of 6.5 across six values fixes the total exactly.

   Turn the mean into a total. There will be 6 values with mean 6.5, so the total must be 6 × 6.5 = 39.

   Subtract what is already there. 3 + 11 + 15 + 6 = 35.

   So the two blanks must sum to 39 − 35 = 4.

   List the counting-number pairs summing to 4. Counting numbers start at 1, so: · 1 + 3 = 4 ✓ · 2 + 2 = 4 ✓ · 3 + 1 = 4 — the same pair as the first, just written the other way round · 0 + 4 is not allowed, since 0 is not a counting number

   Check both. · 3, 11, 1, 3, 15, 6 → sum 39, mean 39/6 = 6.5 ✓ · 3, 11, 2, 2, 15, 6 → sum 39, mean 39/6 = 6.5 ✓

   Counting the possibilities. As sets of two values there are 2 answers: {1, 3} and {2, 2}. If the two blanks are treated as distinguishable positions there are 3 ordered fillings: (1, 3), (3, 1) and (2, 2).

   ✦ The two blanks must add to 4, so they are 1 and 3 or 2 and 2 — 2 possibilities as unordered pairs (3 if the order of the two blanks is counted).

5. 54 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 127, Q5

   Check whether each statement is true, justifying with algebra if needed: (i) The average of two even numbers is even. (ii) The average of any two multiples of 5 will be a multiple of 5. (iii) The average of any 5 multiples of 5 will also be a multiple of 5.

   Hint. One counterexample is enough to demolish a claim — try to build one before trying to prove anything.

   (i) The average of two even numbers is even — FALSE. Counterexample: 2 and 4 are both even, and their average is (2 + 4)/2 = 3, which is odd.

   What is actually true: two even numbers are 2a and 2b, so their average is (2a + 2b)/2 = a + b. This is a whole number, but it is even only when a and b have the same parity. With 2 and 4 we get a = 1, b = 2 and a + b = 3, odd. With 4 and 8 we get 2 + 4 = 6, even. So the average of two even numbers is always a whole number, but not always an even one.

   (ii) The average of two multiples of 5 is a multiple of 5 — FALSE. Counterexample: 5 and 10 are both multiples of 5, and their average is (5 + 10)/2 = 7.5, which is not even a whole number.

   What is actually true: two multiples of 5 are 5a and 5b, so their average is 5(a + b)/2. This is a multiple of 5 only when a + b is even, i.e. when a and b are both odd or both even. For 5 and 10, a = 1 and b = 2, so a + b = 3 is odd and the average is not a multiple of 5. For 5 and 15 we get 5 × 4/2 = 10 ✓.

   (iii) The average of any 5 multiples of 5 is a multiple of 5 — FALSE. Counterexample: 5, 5, 5, 5 and 10 are all multiples of 5, and their average is (5 + 5 + 5 + 5 + 10)/5 = 30/5 = 6, which is not a multiple of 5.

   What is actually true: five multiples of 5 are 5a, 5b, 5c, 5d, 5e, so their average is 5(a + b + c + d + e)/5 = a + b + c + d + e, always a whole number but a multiple of 5 only when a + b + c + d + e happens to be. In the counterexample the bracket is 1 + 1 + 1 + 1 + 2 = 6, giving 6.

   The pattern. Averaging divides, and division can undo the very property being claimed. Multiplying multiples of 5 keeps them multiples of 5; adding them does too; but dividing the sum by the count can strip the factor 5 away.

   ✦ All three statements are false. Counterexamples: (i) 2 and 4 average to 3; (ii) 5 and 10 average to 7.5; (iii) 5, 5, 5, 5, 10 average to 6. In each case the division involved in taking an average can destroy the property being claimed.

6. 65 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 128, Q6

   Two new students joined Sudhakar's class shortly after the class average height was found to be 150.2 cm across 24 students. (i) Which statements are correct: (a) the average will increase because there are 2 new values, (b) the average will remain the same, (c) the new students' heights must be measured, (d) everyone must be measured again? (ii) The new heights are 149 cm and 152 cm — will the average stay the same, increase, decrease, or is there insufficient information? (iii) What about the median?

   Hint. The old total is already known from the old average — that is what decides how much extra measuring is needed.

   (i) Which statements are correct.

   (a) The average will increase because there are 2 new values — WRONG. Adding values does not push the mean in any particular direction; what matters is how tall the newcomers are compared with 150.2 cm. Two students of 140 cm would drag the average down.

   (b) The average will remain the same — WRONG. It stays the same only in the special case where the two new heights average exactly 150.2 cm. There is no reason to expect that.

   (c) The heights of the new students have to be measured — CORRECT. The new mean is (old total + two new heights)/26, and the two new heights are the only pieces of information missing.

   (d) Everyone in the class has to be measured again — WRONG. The old total is already available: 24 × 150.2 = 3604.8 cm. An average always carries the total inside it, so re-measuring 24 students would be pointless work.

   (ii) With heights 149 cm and 152 cm.

   New total = 3604.8 + 149 + 152 = 3905.8 cm, over 26 students:

   New average = 3905.8/26 = 150.223 cm

   So the average increases — option (b), though only by about 0.02 cm.

   Why it increases, seen without dividing: 149 is 1.2 cm below the old mean and 152 is 1.8 cm above it, so the two newcomers bring a net surplus of 0.6 cm. Spread over 26 students that raises the mean by 0.6/26 ≈ 0.023 cm. Equivalently, the newcomers' own average is (149 + 152)/2 = 150.5 cm, which is above 150.2, so they lift the class average.

   (iii) The median. The correct answer is (d) the information is not sufficient.

   Knowing the mean tells us nothing about where the middle student stands. The old median could be 145 cm or 155 cm — both are consistent with a mean of 150.2 — and adding two students at 149 and 152 would move it differently in each case. Without the individual heights, or at least the old median and the values around it, the new median cannot be determined.

   ✦ (i) Only (c) is correct — the new students must be measured, but nobody else need be, since the old total 24 × 150.2 = 3604.8 cm is already known. (ii) (b) The average will increase, to 3905.8/26 ≈ 150.22 cm, because the two newcomers average 150.5 cm, which is above 150.2. (iii) (d) Insufficient information — the mean says nothing about where the middle value lies.

7. 74 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 128, Q7

   A dot plot shows 25 values: 14 (2 dots), 15 (2), 16 (3), 17 (5), 18 (4), 19 (4), 20 (3), 21 (1), 22 (0), 23 (1). Is 17 the average of this data? Share the method you used.

   Hint. Test 17 as a balance point rather than computing the mean from scratch.

   The balance test. The mean is the point where the total distance to the values on its left equals the total distance to the values on its right. So instead of adding all 25 numbers, measure every dot from 17 and see whether the two sides cancel.

   Distances below 17: · 14 is 3 below, twice → 6 · 15 is 2 below, twice → 4 · 16 is 1 below, three times → 3 · Total shortfall = 6 + 4 + 3 = 13

   Distances above 17: · 18 is 1 above, four times → 4 · 19 is 2 above, four times → 8 · 20 is 3 above, three times → 9 · 21 is 4 above, once → 4 · 23 is 6 above, once → 6 · Total surplus = 4 + 8 + 9 + 4 + 6 = 31

   (The five dots at 17 contribute nothing, since they sit on the proposed balance point.)

   The verdict. The right side is heavier by 31 − 13 = +18, so the plank tips right and 17 is not the average. The balance point must lie to the right of 17.

   Where exactly? Spreading the surplus of 18 over all 25 values shifts the balance point by 18/25 = 0.72, so

   mean = 17 + 18/25 = 17.72

   Cross-check by the ordinary method. Total = (14×2) + (15×2) + (16×3) + (17×5) + (18×4) + (19×4) + (20×3) + (21×1) + (23×1) = 28 + 30 + 48 + 85 + 72 + 76 + 60 + 21 + 23 = 443, and 443/25 = 17.72 ✓

   The balance method is quicker here because the deviations are small single-digit numbers, while the raw values are all in the teens and twenties.

   ✦ No, 17 is not the average. The deviations from 17 sum to +18 rather than 0, so the balance point lies to the right; the true mean is 17 + 18/25 = 17.72 (confirmed as 443/25). The median, by contrast, is close to it at 18.

8. 84 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 128-129, Q8

   A group's weights were measured monthly. Last month the mean weight was 65.3 kg and the median 67 kg. This month one person lost 2 kg and two people gained 1 kg each. What can be said about the change in mean weight and median weight?

   Hint. Work out the change in the group's total weight, then ask whether you know enough about the middle of the list.

   The mean. The mean depends only on the total weight and the number of people, and the number of people has not changed.

   Change in total = −2 (one person lost 2 kg) + 1 + 1 (two people gained 1 kg each) = 0 kg

   The total is unchanged and the count is unchanged, so the mean is unchanged at 65.3 kg. It does not matter how many people are in the group, nor which people changed — the losses and gains cancel exactly.

   The median. Here the answer is different: the median cannot be determined.

   The median depends on which values are where in the sorted order, and we are told nothing about which members changed. Three cases show how differently things can go — take a small group of five with weights 60, 63, 67, 70, 74 (median 67): · If the light end changes: 58, 64, 67, 71, 74 → the middle value is untouched, median still 67. · If the middle person is the one who lost 2 kg: 60, 63, 65, 71, 74 → median falls to 65. · If the middle person is one of those who gained: 58, 63, 68, 71, 74 → median rises to 68.

   All three are consistent with everything we were told, and the median comes out lower, higher and unchanged. So while the mean is pinned down exactly, the median is not.

   The lesson. The mean responds only to the total, so changes that cancel leave it alone; the median responds to position in the sorted order, so it needs to be known which individuals moved and by how much.

   ✦ The mean stays exactly the same at 65.3 kg, because the group's total weight changes by −2 + 1 + 1 = 0 and nobody joins or leaves. The median cannot be determined — it could rise, fall or stay at 67 depending on which members of the group lost or gained, and that information is not given.

9. 96 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 129, Q9

   The retail price of iodised salt (₹) each January is given for six states from 2016 to 2025 — Andaman & Nicobar 16 → 20.99, Assam 6 → 12.35, Gujarat 16.5 → 19.2, Mizoram 20 → 29.8, Uttar Pradesh 16.15 → 24.81, West Bengal 9.47 → 23.99. (i) Choose three states and present them on a line graph. (ii) What do you find interesting? (iii) Compare the price variation in Gujarat and Uttar Pradesh. (iv) In which state has the price increased the most from 2016 to 2025? (v) What would you explore further?

   Hint. For (iv), decide first whether 'most' means the largest rupee rise or the largest proportional rise — then check whether the two answers agree.

   (i) Choosing three states for a line graph. A good choice is Assam, Uttar Pradesh and Mizoram — the cheapest, a middling state and the dearest — so that the lines are well separated and the story is visible. Put years 2016 to 2025 along the horizontal axis and price in ₹ up the vertical axis. Since prices run from ₹6 to ₹29.80, a scale of 1 cm = ₹5 from 0 to 30 works; label each line and use different markers as well as different colours.

   (ii) What is interesting. · Every state ends higher than it started, but by wildly different amounts. · Mizoram is the dearest throughout — starting at ₹20 when Assam was at ₹6, and still dearest in 2025 at ₹29.80. Being landlocked and far from the salt-producing coast is the likely reason. · Assam is the cheapest throughout, and its price sits at a flat ₹12 for the four years 2017 to 2020, then wobbles. · Mizoram's prices are suspiciously round — 20, 20, 22, 22, 20, 22, 25 — which suggests reported or controlled prices rather than finely measured ones. · West Bengal has the most dramatic climb, more than doubling from ₹9.47 to ₹23.99 and overtaking Andaman & Nicobar, Assam and Gujarat along the way. · There is a common upward push after 2020, visible in almost every state.

   (iii) Gujarat versus Uttar Pradesh.

   	2016	2020	2025	Total rise	% rise
   Gujarat	16.50	13.00	19.20	₹2.70	16.4%
   Uttar Pradesh	16.15	18.96	24.81	₹8.66	53.6%

   The two started within 35 paise of each other in 2016, and then went in different directions. Gujarat barely moved — it even fell to ₹13 in 2020 and stayed near ₹14–15 for most of the decade before jumping in the final year, so it varies within a narrow ₹13 to ₹19.20 band. Uttar Pradesh rose steadily and much further, from ₹16.15 to a peak of ₹26.90 in 2024. Gujarat is a major salt-producing state, which is a plausible reason its prices stayed low and steady.

   (iv) Which state increased the most. This depends on what 'most' means, so check both.

   State	2016	2025	Rise (₹)	Rise (%)
   A & N Islands	16.00	20.99	4.99	31.2%
   Assam	6.00	12.35	6.35	105.8%
   Gujarat	16.50	19.20	2.70	16.4%
   Mizoram	20.00	29.80	9.80	49.0%
   Uttar Pradesh	16.15	24.81	8.66	53.6%
   West Bengal	9.47	23.99	14.52	153.3%

   West Bengal wins on both counts — the biggest rupee rise (₹14.52) and the biggest proportional rise (153.3%, more than two and a half times its 2016 price). It is worth noticing that the two measures need not agree: Assam's rise of ₹6.35 is smaller than Mizoram's ₹9.80 in rupees, yet larger in percentage terms (105.8% against 49.0%), because Assam started from a much lower base.

   (v) Questions to explore further. Why is Mizoram consistently the dearest and Assam the cheapest — is it distance from the salt pans, or transport cost? What happened around 2020–2021 to push prices up almost everywhere? Why did Gujarat's price fall between 2016 and 2020? How do these rises compare with general food inflation over the same decade — is salt getting dearer faster or slower than other essentials? And what were wages doing over the same period, since that is what decides whether the rise actually hurt anyone?

   ✦ (iv) West Bengal shows the largest increase from 2016 to 2025 — up ₹14.52, from ₹9.47 to ₹23.99, a rise of 153.3% — and it leads on both the absolute and the proportional measure. (iii) Gujarat's price is almost flat (₹2.70, 16.4%) while Uttar Pradesh rose more than three times as much (₹8.66, 53.6%), even though the two states began within 35 paise of each other.

10. 104 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 130, Q10

    Two line graphs show the share of households using electricity or kerosene as their primary lighting source, rural and urban, from 1983 to 2023. Which statements are valid? (i) In 1983 the majority in rural areas used kerosene while the majority in urban areas used electricity. (ii) Kerosene use has decreased over time in both rural and urban areas. (iii) In 2000, 10% of urban households used electricity as a primary lighting source. (iv) In 2023 there were no power cuts.

    Hint. Read each line at the year in question — and check whether the graph is even about the thing being claimed.

    Reading the graphs at 1983. · Rural: kerosene ≈ 83%, electricity ≈ 15% · Urban: electricity ≈ 64%, kerosene ≈ 35%

    (i) VALID. In 1983, 83% of rural households used kerosene, which is a clear majority, and 64% of urban households used electricity, also a majority. Both halves of the statement check out against the graphs.

    (ii) VALID. Both kerosene lines fall steadily throughout. Rural kerosene drops from about 83% in 1983 to nearly 0% by 2023, and urban kerosene from about 35% to nearly 0% over the same period. Neither line rises anywhere, so 'decreased over time in both' is exactly right. The mirror-image electricity lines climb to about 100% in both settings.

    (iii) NOT VALID. In 2000 the urban electricity line stands at about 91%, not 10%. The figure 10% belongs to the urban kerosene line, which is down near 10% around that time. The statement has swapped the two lines — an easy mistake, and exactly why a graph needs its legend read before its numbers.

    (iv) NOT VALID. The graphs say nothing whatever about power cuts. They show only the primary source of energy used for household lighting — which fuel a household mainly relies on. A household can be counted as an electricity user and still face daily outages. This is a claim about data that is simply not on the page, and no reading of the graph can support or refute it.

    The habit worth building. Statements (iii) and (iv) fail in two different ways, and both are common. (iii) misreads which line is which; (iv) asks the graph a question it was never designed to answer. Before accepting any claim about a graph, check both that the right line was read and that the graph actually measures the thing being claimed.

    ✦ Valid: (i) and (ii). Not valid: (iii) and (iv). In 2000 about 91% of urban households used electricity, not 10% — the 10% figure is the urban kerosene share. And the graph reports only the primary lighting source, so it carries no information at all about power cuts.

11. 114 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 130-131, Q11

    A line graph shows the average daily time spent on hobbies and games by urban and rural children against age. (i) How long do urban children aged 10 spend each day? (ii) At what age is the rural average 1.5 hours — (a) 8, (b) 10, (c) 12, (d) 14, (e) 18? (iii) Are these correct: (a) kids aged 15 spend twice as long as kids aged 10; (b) all rural kids aged 15 spend at least 1 hour a day?

    Hint. Read values off the gridlines at 1h and 2h — and for (iii)(b), notice the word 'average' in the graph's title.

    Reading the graph against the 1h and 2h gridlines.

    Age	Urban	Rural
    8	≈ 2 h 23 min	≈ 2 h 41 min
    10	≈ 2 h 06 min	≈ 2 h 25 min
    12	≈ 1 h 46 min	≈ 2 h 04 min
    15	≈ 1 h 06 min	≈ 1 h 19 min
    18	≈ 33 min	≈ 36 min
    20	≈ 20 min	≈ 18 min

    Both curves fall steeply through the teenage years; rural children spend a little longer than urban children up to about age 19, after which the two lines merge.

    (i) Urban children aged 10. The blue (Urban) curve at age 10 sits just above the 2h gridline, so they spend about 2 hours a day (measured at roughly 2 h 06 min).

    (ii) When is the rural average 1.5 hours? Halfway between the 1h and 2h gridlines, the orange (Rural) curve is crossed at age ≈ 14.3. Of the options offered, the nearest is (d) 14 years. (The urban curve reaches 1.5 hours earlier, at about age 13.)

    (iii)(a) Kids aged 15 spend twice as long as kids aged 10 — INCORRECT, and backwards. At 15 the figures are about 1 h 06 min (urban) and 1 h 19 min (rural); at 10 they are about 2 h 06 min and 2 h 25 min. So 15-year-olds spend roughly half as long as 10-year-olds, not twice as long. The curve is falling, so the older group must be lower.

    (iii)(b) All rural kids aged 15 spend at least 1 hour a day — INCORRECT. The graph's title is 'Average Daily Time Spent', so the 1 h 19 min figure is an average across all rural 15-year-olds, not a guarantee about each one. An average of 1 h 19 min is perfectly consistent with many children spending 20 minutes while others spend three hours. A statement about every individual can never be justified from an average alone — the same trap as saying every family has 5.22 members because that is the class average.

    ✦ (i) About 2 hours a day. (ii) (d) 14 years (measured at age ≈ 14.3). (iii)(a) Incorrect — 15-year-olds spend about half as long as 10-year-olds, not twice as long; (b) Incorrect — the graph plots an average, which says nothing about what any individual child does.

12. 124 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 131, Q12

    Individual project: make your own activity strip for different days of the week. (i) Do you eat and sleep at regular times every day, and how long do you spend outdoors? (ii) Calculate the average time spent per activity and represent this average day using a strip. (iii) Track an adult at home and compare.

    Hint. Use the same 48-box format as Manoj, so the results can be compared with his.

    Setting up. Draw a strip of 48 boxes, each box standing for 30 minutes from midnight to midnight, and fix a colour key before you start — for example blue = sleeping, green = eating, purple = washing/exercise, yellow = school and study, orange = friends/hobbies/family, grey = travelling. Fill in one strip per day for at least three days, ideally covering both a school day and a holiday. Record as you go: filling a strip in from memory at bedtime is where the errors creep in.

    (i) Questions to answer from your own strips. Compare the position of the green boxes across days — do your meals fall in the same boxes every day, or do they wander? Do the long blue blocks start and end at the same time? Add up any boxes spent outdoors; if 'outdoors' is not one of your six colours, mark those boxes with a dot so they can be counted separately.

    (ii) Building the average day. For each activity, count the boxes across all the days you tracked and divide by the number of days:

    average boxes per day = (total boxes for that activity)/(number of days)

    Round each result to the nearest whole box, then check that the rounded figures still add to 48 — if they add to 47 or 49, adjust the activity whose value was closest to a halfway point. Now colour a fresh strip using those totals to get your 'average day'. Note that this strip shows how much time each activity took, not when, since the timings differ from day to day.

    Worked illustration: if over 3 days you slept 18, 20 and 19 boxes, the average is (18 + 20 + 19)/3 = 19 boxes = 9.5 hours.

    (iii) Comparing with an adult. Track a parent, grandparent or other adult at home for the same days and build their average strip too. Typical things that show up: adults usually have fewer sleep boxes than children (the sleep graph on page 127 predicts about 8 hours against a child's 9.5); adults have a large block of work where the child has school; and children usually have more orange free-time boxes. Compare the totals activity by activity, and put the two strips side by side so the differences are visible at a glance.

    A useful check on every strip. Each strip must contain exactly 48 boxes, so the six activity totals must add to 48, i.e. 24 hours. If they do not, a box has been missed or double-counted.

    ✦ Use a 48-box strip, one box per 30 minutes, with a fixed colour key. Average each activity across the days tracked — average boxes = total boxes ÷ number of days — and check the averages still sum to 48 before colouring the 'average day' strip. Comparing with an adult's strip usually shows the adult sleeping about 1.5 hours less and having work where the child has school and free time. This is a project question; the answer depends on your own recorded data.

13. 134 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 131, Q13

    Small group project (3–4 members). Do at least one of: (i) track the daily sleep time of all family members for a week, represent it on strips, pool the group's data and find the average and median sleep time of children, adults and the elderly; (ii) collect the daily timings of different Grade 8 schools, including class and break time, and analyse the data.

    Hint. Decide on the exact definitions and the recording format before anyone starts collecting.

    (i) The sleep-time study.

    Collecting. Agree the definition first — the question says daily sleep time includes night sleep, naps and any sleep during the day — so that everyone records the same thing. Each member tracks every family member for 7 days, noting sleep to the nearest half hour, and records it as it happens rather than from memory.

    (a) Representing it on strips. Use the 48-box strip from Manoj's example, one strip per person per day, colouring the sleep boxes. Laying a person's seven strips one above the other shows at a glance whether their sleep is regular or ragged.

    (b) Pooling and calculating. Combine everyone's data and sort each person into children, adults or elderly (agree the cut-offs in advance — for instance under 18, 18 to 59, 60 and above — since different groups using different cut-offs cannot be compared). For each category: · Average = (total sleep hours of everyone in the category over the week)/(number of person-days in that category) · Median = sort all the daily sleep figures for that category and take the middle one, or the average of the two middle ones if the count is even

    Since a group of 4 tracking 4 family members each for 7 days produces 4 × 4 × 7 = 112 readings, use the frequency-table method from page 110 rather than sorting 112 numbers by hand: tally how many readings fall at 6 h, 6.5 h, 7 h and so on, then use running totals to find the middle position.

    (c) Findings to look for. Compare the three categories against the chapter's graph on page 127, which predicts about 9.5 hours at age 6 falling to about 8 hours between 30 and 50 and rising again after 50. Also compare average with median in each category: if one person sleeps very little, the average will sit below the median, and the median will describe the group better. Note weekday-versus-weekend differences too — these often turn out to be larger than the differences between age groups.

    (ii) The school-timings study.

    Collecting. Ask neighbours, relatives and friends about Grade 8 timings at schools anywhere in the country. For each school record: start time, end time, total break time, and hence class time = (end − start) − break. Manoj's school is the worked example: 9:30 am to 4:30 pm is 7 hours including breaks.

    Analysing. Tabulate the schools, then compute the average and median of the total day length and of the class time. Draw a dot plot of total hours — with one dot per school it shows the spread instantly and makes any unusual school obvious. Points worth checking: how much do the schools vary? Do schools that start earlier also finish earlier, or do they simply run longer? Is there a pattern by region, or between government and private schools?

    Presenting. A short table plus one dot plot, with two or three sentences of interpretation, communicates more than a page of raw numbers.

    ✦ For (i), fix a common definition of sleep time and common age cut-offs before collecting; represent each person-day on a 48-box strip, then pool the readings and use a frequency table with running totals to get the average and median for children, adults and the elderly. For (ii), record start time, end time and break time for each school, derive class time, and summarise with a table, a dot plot and the average and median day length. This is a project question; the answers depend on the data your group collects.

14. 145 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 131-132, Q14

    Graphs show sunrise and sunset times across the year at four places in India — Kibithu, Ghuar Moti, Srinagar and Kanyakumari. Which lines are sunrise and which are sunset? (i) At which place does the sun rise earliest in January, and what is the approximate day length there in January? (ii) Which place has the longest day length over the year? (iii) Share your observations.

    Hint. Day length is the vertical gap between a place's two lines at that month.

    Which lines are which. Each place has two curves. The lower band, between 04:00 and 08:00, is sunrise; the upper band, between 16:00 and 20:00, is sunset. Nothing else is possible — the sun cannot rise at 7 pm.

    Reading the four places (measured against the gridlines):

    	Jan sunrise	Jan sunset	Jan day length	Longest day (June)	Shortest day (Dec)
    Kibithu (Arunachal, far east)	05:57	16:28	10 h 31 min	13 h 56 min	10 h 22 min
    Ghuar Moti (Gujarat, far west)	07:42	18:32	10 h 50 min	13 h 35 min	10 h 41 min
    Srinagar (far north)	07:38	17:46	10 h 08 min	14 h 25 min	9 h 54 min
    Kanyakumari (far south)	06:39	18:21	11 h 42 min	12 h 36 min	11 h 39 min

    (i) Earliest sunrise in January: Kibithu, at about 06:00, well over an hour and a half before Ghuar Moti's 07:42. Its day length in January is 16:28 − 05:57 ≈ 10½ hours.

    Why Kibithu? It is the easternmost inhabited place in India, and the whole country runs on a single clock (IST). The sun reaches the east first, so Kibithu's clock shows an early sunrise — and, for the same reason, an early sunset at 16:28, barely past 4:30 in the afternoon.

    (ii) Longest day length over the year: Srinagar, reaching about 14 h 25 min in June — ahead of Kibithu's 13 h 56 min, Ghuar Moti's 13 h 35 min and Kanyakumari's 12 h 36 min.

    (iii) Observations.

    · Latitude controls the swing. Srinagar, the farthest north, has both the longest day (14 h 25 min in June) and the shortest (9 h 54 min in December) — a swing of about 4½ hours. Kanyakumari, close to the equator, swings by barely 1 hour (11 h 39 min to 12 h 36 min), so its days are nearly the same length all year. The further from the equator, the more dramatic the seasons.

    · Longitude controls the clock time. Kibithu and Ghuar Moti have almost the same day lengths month by month (10 h 31 min against 10 h 50 min in January), yet their sunrise times differ by about 1 h 45 min, because India spans nearly 30 degrees of longitude but keeps one time zone. In Kibithu the working day effectively begins and ends in a different part of the daylight than in Gujarat.

    · All four curves turn at the same times of year — sunrise earliest and sunset latest around June, the reverse around December — because they are all in the northern hemisphere and share the same solstices.

    · Sunrise and sunset are not symmetric about noon. Kanyakumari's January day runs 06:39 to 18:21, centred at 12:30, not 12:00 — a consequence of longitude and the equation of time.

    Curious to find out: why is the earliest sunset not on the same date as the latest sunrise? What would these graphs look like for a place in the southern hemisphere — would they be flipped? And what happens above the Arctic Circle, where the two curves would have to meet?

    ✦ The lower curves are sunrise and the upper curves are sunset. (i) Kibithu has the earliest January sunrise at about 06:00, with a January day length of about 10½ hours. (ii) Srinagar has the longest day of the year, about 14 h 25 min in June. (iii) Day length varies most far from the equator (Srinagar swings 4½ hours, Kanyakumari only 1 hour), while Kibithu's and Ghuar Moti's clock times differ by nearly 1 h 45 min because India keeps a single time zone across a very wide country.

15. 155 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 132, Q15

    A graph shows moonrise and moonset times over one month. (i) On what dates were amavasya (new moon) and purnima (full moon) in this month? (ii) What do you notice and what do you wonder?

    Hint. At full moon the moon rises about when the sun sets; at new moon it rises about when the sun rises.

    The two rules that decode the graph. · At purnima (full moon) the moon is opposite the sun in the sky, so it rises at about sunset (≈ 18:00) and sets at about sunrise the next morning. · At amavasya (new moon) the moon is in the same direction as the sun, so it rises at about sunrise (≈ 06:00) and sets at about sunset — which is why it cannot be seen at all.

    Reading the graph. Both series climb steadily and then drop off the top of the chart and restart from the bottom, because a rise time past midnight becomes an early time on the next date. · The moonrise line runs from about 08:30 on the 1st up to midnight around the 20th, then restarts near 00:30 on the 22nd and climbs to about 08:30 by the 31st. · The moonset line runs from about 19:00 on the 1st to midnight around the 5th, then restarts near 00:00 on the 7th and climbs to about 20:00 by the 31st.

    (i) The two dates. · Purnima: the moonrise line crosses 18:00 at about the 13th–14th. On the 14th the graph gives moonrise 18:28 and moonset 07:34 — the moon up nearly all night, which is exactly what a full moon does. So purnima was around the 14th. · Amavasya: the moonrise line crosses 06:00 at about the 28th, and the moonset line crosses 18:00 at about the 29th — the moon rising and setting with the sun. So amavasya was around the 28th–29th.

    The two dates are about 14 to 15 days apart, which is the built-in check: purnima and amavasya are always half a lunar month apart.

    (ii) What to notice.

    · The moon rises about 50 minutes later each day. Moonrise moves from 08:30 on the 1st to nearly midnight on the 20th — about 15½ hours in 19 days, roughly 49 minutes per day. The moonset line climbs at the same rate. · That daily lag explains the month. At 50 minutes a day, moonrise works its way right round the 24-hour clock in 24 × 60 ÷ 50 ≈ 29 days — which is the length of a lunar month, and why our months are roughly 29 or 30 days long. · Unlike the sun, the moon does not keep to the night. For roughly half of every month it is above the horizon in broad daylight, which is why the moon is often visible in the afternoon sky. · The lines are straight, not curved. The sunrise/sunset curves in Q14 bend with the seasons, but moonrise marches along at a near-constant slope, because it is driven by the moon's orbit rather than by the earth's tilt.

    What to wonder. Why is the lag about 50 minutes and not some other amount? Do the phases repeat on the same dates next month, or do they drift? How do lunar calendars — and festivals such as Diwali on amavasya and Holi and Guru Purnima on purnima — keep in step with the solar year? And can this graph be used to predict which nights will be dark enough for stargazing?

    ✦ Purnima (full moon) was around the 14th, where moonrise ≈ 18:00 and moonset ≈ 06:00 the next morning; amavasya (new moon) was around the 28th–29th, where moonrise ≈ 06:00 and moonset ≈ 18:00 — about 14 to 15 days apart, as they must be. The graph's main lesson is that moonrise comes about 50 minutes later each day, which is exactly why a lunar month is about 29 days long.

Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp205.pdf), Chapter 5 'Tales by Dots and Lines', pages 103-133. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every numeric answer was derived from first principles and independently recomputed in Python. MEASURED OFF THE PRINTED FIGURES at 300-1200 dpi: the page-113 dot plot reads 4, 7, 8, 8, 9, 9, 9, 9, 9, 11 (ten dots, so the missing eleventh is 16); the three page-114 album dot plots read A = 5, 5, 5.25, 5.5, 5.75, 6, 6.5 (mean 39/7 = 5.5714), B = 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5 (mean 2.406) and C = 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5 (mean 3.906), so A is the 5.57 album; the page-115 cycle dot plot reads 0:3, 1:1, 2:4, 3:7, 4:7, 5:5, 6:4, 7:6, 8:3, 9:0, 10:2 (N = 42, sum 193, mean 4.595, median 4 — and the book's own hint that four students rode twice confirms the reading); the page-128 dot plot reads 14:2, 15:2, 16:3, 17:5, 18:4, 19:4, 20:3, 21:1, 22:0, 23:1 (N = 25, sum 443, mean 17.72); the page-122 New Delhi rainfall line reads 1.3, 1.5, 1.5, 1.1, 1.5, 3.8, 9.7, 9.7, 4.0, 1.0, 0.4, 1.0 days, total about 37, which makes New Delhi the least-rainy of the four cities and Port Blair the most at 125.8; the page-123 births line graph was found to span April 2017 to March 2020 (36 monthly points) with July 2017 about 1.77 M, January 1.67/1.75/1.77 M in 2018/19/20 and a 2019 total of about 21.4 M; the page-124 Wheat-vs-Rice infographic was read state by state and Karnataka's hidden shade was matched against the colour bar (calibrated exactly on Kerala +79 and Chhattisgarh +80) to about +68; the page-125 activity strips were decoded box by box across all 48 boxes of all three strips, giving Friday/Saturday/Sunday and an identical 10.5 hours of sleep and 1.5 hours of eating on each day; the page-130 hobbies line graph gives urban age 10 about 2 h 06 min and rural 1.5 h at age about 14.3, so option (d); the page-132 sunrise/sunset charts give Kibithu the earliest January sunrise (05:57, day length 10 h 31 min) and Srinagar the longest day of the year (14 h 25 min in June); and the page-132 moon chart gives purnima about the 14th and amavasya about the 28th-29th with a measured daily lag of 49 minutes. TWO SLIPS IN THE PRINTED BOOK ARE FLAGGED: page 125 says Manoj recorded 'five types of activities' and then lists six (the strips do use six colours), and the page-122 Figure it Out numbers two different questions as '2'.. Questions are referenced from the NCERT textbook for identification.